高访问员工
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题目描述
You are given a 2D 0-indexed array of strings, access_times, with size n. For each i where 0 <= i <= n - 1, access_times[i][0] represents the name of an employee, and access_times[i][1] represents the access time of that employee. All entries in access_times are within the same day.
The access time is represented as four digits using a 24-hour time format, for example, "0800" or "2250".
An employee is said to be high-access if he has accessed the system three or more times within a one-hour period.
Times with exactly one hour of difference are not considered part of the same one-hour period. For example, "0815" and "0915" are not part of the same one-hour period.
Access times at the start and end of the day are not counted within the same one-hour period. For example, "0005" and "2350" are not part of the same one-hour period.
Return a list that contains the names of high-access employees with any order you want.
Example 1:
Input: access_times = [["a","0549"],["b","0457"],["a","0532"],["a","0621"],["b","0540"]] Output: ["a"] Explanation: "a" has three access times in the one-hour period of [05:32, 06:31] which are 05:32, 05:49, and 06:21. But "b" does not have more than two access times at all. So the answer is ["a"].
Example 2:
Input: access_times = [["d","0002"],["c","0808"],["c","0829"],["e","0215"],["d","1508"],["d","1444"],["d","1410"],["c","0809"]] Output: ["c","d"] Explanation: "c" has three access times in the one-hour period of [08:08, 09:07] which are 08:08, 08:09, and 08:29. "d" has also three access times in the one-hour period of [14:10, 15:09] which are 14:10, 14:44, and 15:08. However, "e" has just one access time, so it can not be in the answer and the final answer is ["c","d"].
Example 3:
Input: access_times = [["cd","1025"],["ab","1025"],["cd","1046"],["cd","1055"],["ab","1124"],["ab","1120"]] Output: ["ab","cd"] Explanation: "ab" has three access times in the one-hour period of [10:25, 11:24] which are 10:25, 11:20, and 11:24. "cd" has also three access times in the one-hour period of [10:25, 11:24] which are 10:25, 10:46, and 10:55. So the answer is ["ab","cd"].
Constraints:
1 <= access_times.length <= 100access_times[i].length == 21 <= access_times[i][0].length <= 10access_times[i][0]consists only of English small letters.access_times[i][1].length == 4access_times[i][1]is in 24-hour time format.access_times[i][1]consists only of'0'to'9'.
代码结果
运行时间: 42 ms, 内存: 16.0 MB
/*
* 题目思路：
* 1. 使用 Java Stream API 对数据进行处理。
* 2. 将访问时间转换为分钟并分组到相应的员工。
* 3. 对每个员工的时间列表进行排序并检查是否存在连续的三次访问是在同一小时内。
* 4. 如果存在这样的情况，将员工加入到高访问员工列表中。
*/
import java.util.*;
import java.util.stream.*;
public class HighAccessEmployeesStream {
public List<String> getHighAccessEmployees(String[][] access_times) {
// 将访问时间转换为分钟并分组到相应的员工
Map<String, List<Integer>> accessMap = Arrays.stream(access_times)
.collect(Collectors.groupingBy(
access -> access[0],
Collectors.mapping(access -> Integer.parseInt(access[1].substring(0, 2)) * 60 + Integer.parseInt(access[1].substring(2)),
Collectors.toList())));
// 检查每个员工是否为高访问员工
return accessMap.entrySet().stream()
.filter(entry -> {
List<Integer> times = entry.getValue();
Collections.sort(times);
for (int i = 0; i <= times.size() - 3; i++) {
if (times.get(i + 2) - times.get(i) < 60) {
return true;
}
}
return false;
})
.map(Map.Entry::getKey)
.collect(Collectors.toList());
}
}
解释
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